Question

determine o Tg x, onde cós x= 1/5​

Answer 1

Entre as identidades trigonométricas, encontramos $\sin^2x + \cos^2x = 1$, que poderemos usar para encontrar o seno de x:

$\sin^2x + (\frac{1}{5})^2 = 1\\\\\sin^2x + \frac{1}{25} = 1\\\\\sin^2x = 1 - \frac{1}{25}\\\\\sin^2x = \frac{24}{25}\\\\\sin x = \pm\sqrt{\frac{24}{25}}\\\\\sin x = \pm\frac{2\sqrt{6}}{5}$

Então, como a tangente é a razão entre o seno e o cosseno, temos:

$\tan x = \pm\frac{\sin x}{\cos x}\\\\\tan x = \pm\frac{2\sqrt6}{5} \div \frac{1}{5} = \pm\frac{2\sqrt6}{5} \times 5\\\\\tan x = \pm2\sqrt6$

Logo, os valores possíveis de tangente são:

$\tan x = 2\sqrt6\\\\\tan x = -2\sqrt6$

Answer 2

$\displaystyle \text {sen}^2(\text x)+\text{cos}^2(\text x)=1 \\\\ \text{sen}^2(\text x) = 1-\text{cos}^2(\text x) \\\\ \text{sen(x)}=\pm\sqrt{1-\text{cos}^2(\text x)} \\\\ \underline{\text{sabemos que a tangente {\'e} igual a}}: \\\\ \text{tg}(\text x)=\frac{\text{sen(x)}}{\text{cos(x)}} \\\\\\ \text{tg(x)}=\frac{\pm\sqrt{1-\text{cos}^2(\text x)}}{\text{cos(x)}} \\\\\\ \underline{\text{substituindo o calor de cos(x)} }:$

$\displaystyle \\\\ \text{tg(x)}=\pm\frac{\sqrt{\displaystyle 1-\frac{1}{5^2}}}{\displaystyle \frac{1}{5}} \\\\\\ \text{tg(x)}=\pm\frac{\displaystyle \sqrt{1-\frac{1}{25}}}{\displaystyle \frac{1}{5}} \\\\\\ \text{tg(x)}= \pm\frac{\displaystyle \frac{\sqrt{25-1}}{5}}{\displaystyle \frac{1}{5}} \\\\ \text{tg(x)}=\pm\sqrt{24} = \pm\sqrt{4.6} = \pm2\sqrt{6} \\\\ \underline{\text{Portanto}}:\\\\\ \huge\boxed{\ \text{tg(x)}=2\sqrt{6}\ } \\\\\\ \text{ou} \\\\\ \huge\boxed{\ \text{tg(x)}=-2\sqrt{6}\ }$